I am going over Durret's probability textbook and one of the proofs uses the inequality
$$\sum_{k=n+1}^\infty \frac{1}{k^2\log k} \leq \frac{1}{n\log n}$$
But, as with much of the textbook, gives no explanation. I was thinking you could take one of the $k$'s out front to and then integrate $1/k\log k$ but that doesn't seem to be of much help.
How can I show this to be true (rather than just take Durret's word for it)?
Use an integral estimate for the series: $$ \sum_{k=n+1}^{+\infty} \frac{1}{k^2\log k} \leq \int_n^{+\infty} \frac{dx}{x^2\log x} = \int_{\log n}^{+\infty} \frac{dt}{te^t} \leq \frac{1}{\log n} \int_{\log n}^{+\infty} e^{-t}\,dt=\frac{1}{n\log n}. $$